When Itô’s Isometry Applies to Stochastic Integrals

Summary

The answer clarifies that the familiar second-moment identity for a stochastic integral applies when the integrator is a martingale and the integrand satisfies suitable integrability conditions. It warns against applying the identity directly to a process that is not a martingale. In the example, the target quantity includes a deterministic drift contribution as well as a Brownian stochastic integral, so its second moment cannot be obtained from the quadratic variation of the full process alone.

The calculation expands the square, uses the zero expectation of the stochastic integral to eliminate the cross term, and then applies Itô’s isometry to the Brownian integral. It evaluates the remaining expectation using the second moment of Brownian motion. This illustrates the distinction between drift and martingale components in a second-moment calculation. The response focuses on this example and does not discuss the broader conditions for square integrability or other forms of stochastic integration.

Key ideas

• Itô’s isometry applies to stochastic integrals with martingale integrators under appropriate integrability conditions.
• A process with drift is not itself a martingale, so its quadratic variation alone does not determine the integral’s second moment.
• Separate drift and stochastic-integral terms before expanding the square.
• The Brownian integral’s second moment can be evaluated by applying the isometry to its integrand.

Tags

# Stochastic calculus: what am I doing wrong?


# Stochastic calculus: what am I doing wrong?












it is just the computation of a second moment but however is creating debate !!... Can someone spot the error?

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/18123

For a martingale $\{M_t \mid t\geq 0\}$ and the stochastic integral \begin{align*} I_t = \int_0^tZ_s dM_s, \end{align*} we have that \begin{align*} E((I_t)^2) = E\bigg( \int_0^tZ_s^2 d\langle M\rangle_s\bigg), \end{align*} where $\langle M\rangle$ is the quadratic variation. That is, the ito's isometry holds for a martingale integrator only.

However, in your question, $\{X_t \mid t\geq 0\}$ is not a martingale, then \begin{align*} E\big( Y_t^2)\big) \neq E\bigg(\int_0^t (1+s)^2 d\langle X\rangle_s \bigg). \end{align*} Instead, since \begin{align*} Y_t^2 &= \bigg(4\int_0^t(1+s)ds + 6\int_0^t(1+s)W_sdW_s \bigg)^2\\ &=16\bigg( \int_0^t(1+s)ds\bigg)^2 + 48 \int_0^t(1+s)ds\int_0^t(1+s)W_sdW_s + 36 \bigg( \int_0^t(1+s)W_sdW_s\bigg)^2, \end{align*} then \begin{align*} E\big( Y_t^2)\big) &= 16\bigg( \int_0^t(1+s)ds\bigg)^2 + 36 E\bigg[\bigg( \int_0^t(1+s)W_sdW_s\bigg)^2\bigg]\\ &=16\bigg( \int_0^t(1+s)ds\bigg)^2 + 36 E\bigg[\bigg( \int_0^t(1+s)^2W_s^2ds\bigg)\bigg]\\ &= 16\bigg( \int_0^t(1+s)ds\bigg)^2 + 36\int_0^t(1+s)^2 s\, ds. \end{align*}

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